Let $$A = \Big\{(a_1,a_2,\dots)\ \Big|\ a_i\ge 0, \sum_{i=1}^\infty a_i=1\Big\},$$ $$v(x)=\sup\left(\bigg\{\sum_{i=1}^\infty a_ib_i\ \bigg|\ (a_i)_{i=1}^\infty,\, (b_i)_{i=1}^\infty \in A,\,\sup\limits_{i\in \mathbf N} a_ib_i =x\bigg\}\right).$$ Certainly $v$ is an increasing function. Is $v(x)$ finite for every $x$? Is it achievable? Does $v(x)\rightarrow 0$ as $x\rightarrow 0$?
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2 Answers
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We have the following inequality:
\begin{align*}\sum_{i =1}^\infty a_i b_i & \le \left(\sum_{i = 1}^\infty a_i^2 b_i\right)^{1/2} \left(\sum_{i = 1}^{\infty} b_i\right)^{1/2} \\ & \le \left(\sum_{i = 1}^\infty a_i a_i b_i\right)^{1/2} \\ & \le \left((\sup_i a_i b_i) \sum_{i =1}^\infty a_i\right)^{1/2}\\ & \le \sqrt{x}.\end{align*}
Hence $\nu(x) \to 0$ as $x \to 0$.
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$\begingroup$ Thank you for the elegant solution. Would you mind I add a very slight alternative? $\endgroup$– HansCommented Oct 16, 2014 at 18:10
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$\begingroup$ Of course not. Would you mind to explain me why you encounter such question? $\endgroup$ Commented Oct 16, 2014 at 20:34
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$\begingroup$ I was actually banging my head: why I did not think of this solution earlier... :-/ For the origin of the problem, you can take a look at my answer to my own question quant.stackexchange.com/questions/14994/…. $\endgroup$– HansCommented Oct 16, 2014 at 21:29
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Inspired by the solution of Yanqi Qiu, here is a more symmetric derivative of it.
\begin{align} \Big( \sum_i a_ib_i\Big)^2 &= \Big(\sum_i (a_ib_i)^\frac{1}{2}a_i^\frac{1}{2}b_i^\frac{1}{2}\Big)^2 \\ &\le \sup\limits_i (a_ib_i)\Big(\sum_i a_i^\frac{1}{2}b_i^\frac{1}{2}\Big)^2 \\ &\le \sup\limits_i (a_ib_i)\sum_i a_i\sum_ib_i. \end{align}